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lorasvet [3.4K]
3 years ago
6

A football coach gives half his players a protein bar to eat before each game; the other players do not receive the protein bar.

After a month of games, he analyzes their playing stats. He concludes the protein bar helped improve their performance on the field. How can the coach test the means to be sure the results are not likely to happen by chance?
Mathematics
1 answer:
timofeeve [1]3 years ago
4 0
To make the sure the results did not happen by chance, the coach can ensure that the protein bars are the only difference between the two groups. For example, he can make sure that both groups are drinking the same amounts of water and getting in the same amount of practice; that way, he will be able to determine that one group is performing better due to the protein bars only. Also, the coach can repeat this experiment multiple times to make sure that the results weren't a random, one-time thing. 
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Answer:

y=-3x-2

Step-by-step explanation:

There is enough information to make a point-slope form equation that which we can convert into slope-intercept form.

Point-slope form is: y-y_1=m(x-x_1)

We are given the slope of -3 and the point of (1,-5).

y-y_1=m(x-x_1)\rightarrow y+5=-3(x-1)

Convert into Slope-Intercept Form:

y+5=-3(x-1)\\y+5-5=-3(x-1)-5\\\boxed{y=-3x-2}

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Is (x - 1) a factor of P(x) = 4x? - 9x2 + 4x+1? If it is, write P(x) as a product of
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Step-by-step explanation:

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Answer:

z=\frac{0.843-0.653}{\sqrt{0.748(1-0.748)(\frac{1}{300}+\frac{1}{300})}}=5.358    

p_v =2*P(Z>5.358) = 4.2x10^{-8}  

Comparing the p value with the significance level given \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can say that we have singificantly differences between the two proportions.  

Step-by-step explanation:

Data given and notation  

X_{1}=253 represent the number with no defects in sample 1

X_{2}=196 represent the number with no defects in sample 1

n_{1}=300 sample 1

n_{2}=300 sample 2

p_{1}=\frac{253}{300}=0.843 represent the proportion of number with no defects in sample 1

p_{2}=\frac{196}{300}=0.653 represent the proportion of number with no defects in sample 2

z would represent the statistic (variable of interest)  

p_v represent the value for the test (variable of interest)  

\alpha=0.05 significance level given

Concepts and formulas to use  

We need to conduct a hypothesis in order to check if is there is a difference in the the two proportions, the system of hypothesis would be:  

Null hypothesis:p_{1} - p_2}=0  

Alternative hypothesis:p_{1} - p_{2} \neq 0  

We need to apply a z test to compare proportions, and the statistic is given by:  

z=\frac{p_{1}-p_{2}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{1}}+\frac{1}{n_{2}})}}   (1)  

Where \hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{253+196}{300+300}=0.748  

z-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.  

Calculate the statistic  

Replacing in formula (1) the values obtained we got this:  

z=\frac{0.843-0.653}{\sqrt{0.748(1-0.748)(\frac{1}{300}+\frac{1}{300})}}=5.358    

Statistical decision

Since is a two sided test the p value would be:  

p_v =2*P(Z>5.358) = 4.2x10^{-8}  

Comparing the p value with the significance level given \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can say that we have singificantly differences between the two proportions.  

5 0
3 years ago
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